I am currently working through Morrison's, Quantum States of Atoms Molecules and Solids, and am having some trouble understanding his derivation of Bloch's theorem of the first form.

We begin with the notion that the Hamiltonian is invariant under transformations for a periodic potential, $V(r)=V(r+T)$, where $T$ is the lattice translation vector, $T=n_1\textbf{a}+n_2\textbf{b}+n_3\textbf{c}$, such that $H(r)=H(r+T)$.

My first question arises here: Why can we assume that the kinetic energy term is invariant under translation?

Regardless, assuming it is invariant under translation for the purpose of this question, we define a translation operator, $\hat{T}$, such that, $\hat{T}f(r)=f(r+T)$, and that $\hat{T}$ can be represented by a function $T(n_1,n_2,n_3)$. Consequently, there are an infinite number translation operators, indexed by $n_1,\,n_2$, and $n_3$. We may hence define a second translation operator, $\hat{T}'$, indexed by $n_1',\,n_2'$, and $n_3'$, such that $\hat{T}\hat{T}'=\hat{T}'\hat{T}$, and therefore, that $[\hat{T},\hat{T'}]=0$.

The book goes on to define $\hat{T}''=\hat{T}\hat{T}'$, corresponding to the translation vector, $T''$, indexed by $n_1+n_1',\,n_2+n_2'$, and $n_3+n_3'$.

Due to (the presumed) translational invariance of the Hamiltonian, we also know that $[H,\hat{T}]=0$, for any $\hat{T}$ (including primed $\hat{T}$s), such that $\hat{T}\psi=c\psi$, $\hat{T}'\psi=c'\psi$, and $\hat{T}''\psi=c''\psi$, where $\psi=\psi_E$ are the energy eigenfunctions. Imposing normalization conditions, we can say that $|c|^2=|c'|^2=|c''|^2=1$. The book at this point goes on to write these constants in the form, $e^{i\beta},\,e^{i\beta'}$, and $e^{i\beta''}$. Because information cannot be lost $\beta,\,\beta'$, and $\beta''$, must be indexed by $n_1,\,n_2,\,n_3$, $n_1',\,n_2',\,n_3'$, and $n_1+n_1',\,n_2+n_2',\,n_3+n_3'$, respectively.

This gives us, $\beta''(n_1+n_1',\,n_2+n_2',\,n_3+n_3')=\beta(n_1,\,n_2,\,n_3)+\beta'(n_1',\,n_2',\,n_3')$. At this point, the book states the following:

The point to be emphasized is that $\beta''$ is the sum of $\beta$ and $\beta'$, and depends on the sums of the corresponding indices, such that for the above equation to hold, $\beta$ must be proportional to $T$, where $T$ is the lattice translation vector, such that $\beta\propto T\to\beta=k\cdot T$, where $k$ is not dependent on the indices.

I am simply not following the jump in logic here.

Why must $\beta=k\cdot T$ for $\beta''(n_1+n_1',\,n_2+n_2',\,n_3+n_3')=\beta(n_1,\,n_2,\,n_3)+\beta'(n_1',\,n_2',\,n_3')$ to hold?


1 Answer 1


Why can we assume that the kinetic energy term is invariant under translation?

Because the kinetic energy only depends on the velocity, not on the location. The velocity does not change under the coordinate transformation, $\frac{d}{dt} (r+T) = \frac{d}{dt} r$.

We may define a second translation operator $\hat T'$ such that $\hat T\hat T'=\hat T'\hat T.$

Maybe you just formulated this in a strange way, but to make sure you didn't misunderstand something: $$\hat T\hat T'=\hat T'\hat T$$ is true for all translation operators $\hat T$ and $\hat T'$.

Due to translational invariance of the Hamiltonian, we also know that $[H,\hat{T}]=0$ for any $\hat{T}$, such that $\hat{T}\psi=c\psi$, $\hat{T}'\psi=c'\psi$, and $\hat{T}''\psi=c''\psi$.

This is wrong! The correct statement is that we can choose an energy eigenbasis such that every basis element $\psi$ is an eigenstate of all translation operators. (But we can easily make linear combinations that are not eigenstates of the translation operators!)

Why must $\beta = k \cdot T$?

Does your book really use that notation? It is horrible in my opinion. For each set of indices $(n_1, n_2, n_3)$, there is one exponent $\beta$. That means there is one function $\beta(n_1, n_2, n_3)$. We can write $\beta' = \beta(n_1', n_2', n_3')$, but there is no function $\beta'(\dots)$. So what we in fact know is $$ \beta(n_1+n_1', n_2+n_2', n_3+n_3') = \beta(n_1, n_2, n_3) + \beta(n_1', n_2', n_3') . $$ $\beta$ is a linear function and therefore $\beta = k \cdot (n_1, n_2, n_3)$. If that's still unclear, consider e.g. $$ \beta(2,1,0) = \beta(1,0,0) + \beta(1,0,0) + \beta(0,1,0) = 2 \beta(1,0,0) + \beta(0,1,0) \equiv 2 k_1 + k_2 .$$

In general, I recommend visiting Wikipedia, they have a very nice (and simpler) write-up of the proof.

  • $\begingroup$ When I said $H\psi=E\psi$, I was in fact referring to $\psi_E$, where $\psi_E$ are the energy eigenfunctions. Thanks for pointing out that it might not be clear to others -- I'll update the question to reflect that fact. Also, just to clarify, by virtue of the fact that $\beta$ is linear, we can say that $\beta=k\cdot T$? $\endgroup$
    – W. Ryan
    Mar 26, 2017 at 13:43
  • $\begingroup$ Yes, I was trying to explain why in the last line :) You could also say, define $k_1 = \beta(1,0,0)$, $k_2 = \beta(0,1,0)$, $k_3=\beta(0,0,1)$. Then it is easy to see that $\beta(n_1,n_2,n_3) = k_1 n_1 + k_2 n_2 + k_3 n_3 = k \cdot T$. (Mathematically, this can be seen as an application of the Riesz representation theorem) $\endgroup$
    – Noiralef
    Mar 26, 2017 at 15:21

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